Problem 34 - Precalculus Mathematics for Calculus

Question

Rewrite the expression as an algebraic expression in \(x\). $$\sin \left(\tan ^{-1} x\right)$$

Step-by-Step Solution

Answer

\( \frac{x}{\sqrt{x^2 + 1}} \)

Step 1: Understand the Problem

The task is to rewrite \( \sin( \tan^{-1}(x) ) \) as an algebraic expression in \( x \). This involves using trigonometric identities to transform the inverse trigonometric function.

Step 2: Set Up the Triangle

Consider \( \theta = \tan^{-1}(x) \). This means \( \tan(\theta) = x \). Recall that \( \tan(\theta) = \frac{{\text{opposite}}}{{\text{adjacent}}} \) in a right triangle, so you can set the opposite side to \( x \) and the adjacent side to \( 1 \).

Step 3: Find the Hypotenuse

Use the Pythagorean theorem: \[ \text{hypotenuse} = \sqrt{x^2 + 1^2} = \sqrt{x^2 + 1} \].

Step 4: Express \(\sin(\theta)\)

The sine of angle \( \theta \) is calculated as \( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{x}{\sqrt{x^2 + 1}} \).

Step 5: Solution

Thus, \( \sin( \tan^{-1}(x) ) = \frac{x}{\sqrt{x^2 + 1}} \).

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions, like \( \tan^{-1} \), allow us to find an angle whose trigonometric ratio is a given value. In the problem, we see \( \tan^{-1}(x) \), also called "arctangent." This is the inverse function of tangent. Therefore, when it says \( \tan^{-1}(x) \), it means finding the angle \( \theta \) such that \( \tan(\theta) = x \).

• These functions are essential for solving problems where the trigonometric ratio is known, but the angle is not.
• The range of \( \tan^{-1}(x) \) is typically from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\).

The use of inverse trigonometric functions connects angles to real-world quantities, ideal for describing cycles, waves, and rotations in physics and engineering. They help transform values into meaningful angles, simplifying complex algebraic expressions.

Pythagorean Theorem

The Pythagorean theorem is a fundamental principle in geometry. It helps us relate the lengths of sides in right-angled triangles. It states that in a right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

In this exercise, we created a triangle as part of solving the trigonometric expression \( \sin( \tan^{-1}(x) ) \). By knowing \( \tan(\theta) = x \), it translates to having a triangle where the opposite side is \( x \) and the adjacent side is \( 1 \).

To find the hypotenuse:\[\text{hypotenuse} = \sqrt{x^2 + 1}\]
This allows us to find the sine of the angle \( \theta \), which ties back to the original problem. By finding the hypotenuse using the Pythagorean theorem, we can graphically solve for other trigonometric functions, simplifying the algebraic expression into \( \frac{x}{\sqrt{x^2 + 1}} \).

• Always ensures accurate calculations for lengths in a triangle.
• Fundamental for converting between different trigonometric ratios.

Algebraic Expressions

Algebraic expressions involve numbers, variables, and arithmetic operations. They are foundational in mathematics, providing a systematic way to represent numbers symbolically. In the exercise, we need to rewrite \( \sin( \tan^{-1}(x) ) \) as an algebraic expression. This involves simplifying trigonometric and inverse trigonometric functions into a rational form.

The expression derived, \( \frac{x}{\sqrt{x^2 + 1}} \), shows the power of algebra in manipulating and simplifying trigonometric identities.

• Algebra helps transform complex expressions into simpler forms by relying on mathematical operations.
• It offers a way to express mathematical ideas without calculating exact numerical values.

Understanding algebraic expressions helps students gain insights into not only solving equations but transforming and simplifying them, which is crucial in various fields of mathematics and science.